Gauss's Law Problems

  • Thread starter zorro
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  • #1
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Homework Statement



1) A point charge is placed at the centre of a spherical Gaussian surface. Is ΦE changed,
a) if the surface is replaced by a cube of the same volume?
b) if the charge is moved off centre in the original sphere, still remaining inside?

2) A point charge q is placed at the centre of a cube of edge a. What is the flux through each of the cube faces? A point charge q is placed at one corner of a cube of edge a. What is the flux through each of the cube faces?

The attempt at solution

1) a) E remains same as the charge is not changed in magnitude. Surface area of the cube is different, Φ=Edscosθ
So ds being different ΦE will change.
b) If we consider the sphere to be our Gaussian surface, the charge enclosed does not change in its magnitude. Hence no change in ΦE

2) ΦE = q/6ε for both the cases as E,ds,q all remain same in both cases.

1) a) and 2) second part is wrong. Please explain.
 

Answers and Replies

  • #2
rl.bhat
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1(a), total flux from a point charge does not depend on the nature of the Gaussian surface. Σdφ = ΣE*ds*cosθ. Here E is not constant at the different points on the surface of the cube.

2. who many small identical cubes can be formed in a larger cube.
 
  • #3
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1. Σdφ = ΣE*ds*cosθ. Here E is not constant at the different points on the surface of the cube.
2. who many small identical cubes can be formed in a larger cube.

1. Do you mean E and ds adjust themselves in such a way that flux remains same?
2. I did not get you.
 
  • #4
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1) Yes. Flux is a measure as how much stuff is coming out, how much is diverging. The surface has changed, but the amount of stuff coming out is the same.

2) You would be right for the first part. For the second part you would find life much easier if you made a larger cube out of the smaller cube so that the charge could be in the center again. Then readjust to get the flux coming out of the small cube. The charge is no longer "inside" the cube, so you'd expect a flux change.
 
  • #5
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2) You would be right for the first part. For the second part you would find life much easier if you made a larger cube out of the smaller cube so that the charge could be in the center again. Then readjust to get the flux coming out of the small cube. The charge is no longer "inside" the cube, so you'd expect a flux change.

I don't understand what is the problem if the charge doesnot remain in the centre? As you said flux is a measure of the 'stuff' coming out - q does not change in magnitude- so 'stuff' coming out doesnot change.
Why is the charge no longer inside? Don't you think the corner of your room is not inside your room?
 
  • #6
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By your analogy, if I went to the point corner of the room, I'd have to go to the wood; no longer inside the room. Geometrically, you can't be inside a figure if you are at an edge that is defining a figure.
 
  • #7
rl.bhat
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1. Do you mean E and ds adjust themselves in such a way that flux remains same?
2. I did not get you.

The point charge emits electric field in all direction.

You can imagine a charge on the corner of a cube as a charge at the center of a large cube made of eight identical cubs. The cube contain six surfaces, each having four squares. Total squares are 24. Total flux is divided into 24 equal parts. Three surfaces of each cube produces the flux. So φ = 3*q/24*εο = q/8εο.
 

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